46 MATH 101A: ALGEBRA I PART B: RINGS AND MODULES10. Jordan canonical formAs an application of the structure theorem for PID’s I explained theJordan canonical form for matrices over the complex numbers. FirstI stated the theorem and then I proved it by reducing it to a moduleover a PID.10.1. statement of the theorem. The theorem is that any n × nmatrix A with coefficients in the complex numbers is conjugate to amatrix in Jordan canonical form. This is defined to be a square matrixwith the eigenvalues of A along the diagonal, 0’s and 1’s on the super-diagonal (right above the diagonal) and zeroes everywhere else. Also,if there is a 1 in the i, i + 1 position then aii= ai+1,i+1. Here is anexample:B−1AB =a 1 0 0 0 00 a 1 0 0 00 0 a 0 0 00 0 0 a 0 00 0 0 0 b 10 0 0 0 0 b.This is a matrix with three Jordan blocksB−1AB =a 1 00 a 10 0 a⊕ (a) ⊕'b 10 b(.On the second day, I used the notation [λ]kto denote the Jordan blockof size k and eigenvalue λ. So, this decomposition would be written:B−1AB = [a]3⊕ [a]1⊕ [b]2.10.2. module over a PID. The key point is something I explainedearlier. Namely, any square matrix over C makes Cninto a moduleover C[T ].If A is an n×n matrix over C then A is an element of the matrix ringMn(C). We let R = C[T ]. This is a PID since C[T] has a Euclideanalgorithm: when you divide one polynomial by another, the remainderhas smaller degree. The module is M = Cnwith the action of the ringgiven by the evaluation mapevA: C[T ] → Mn(C)which sends f(T) to f(A). In other words,f(T )x = f(A)x.MATH 101A: ALGEBRA I PART B: RINGS AND MODULES 47For example, if f(T ) = aT2+ bT + c thenf(T)x = f(A)x = aA2x + bAx + cx.Lemma 10.1. M = Cnis a finitely generated torsion module overR = C[T ].Proof. This uses the “restriction of scalars” idea. Since C[T ] containsthe field C, we can restrict scalars and view M as a vector space overC. Then M is finite dimensional. But C[T ] is infinite dimensionalover C. So, C[T ] cannot be a direct summand of M. So, it is torsion.Also, being finitely generated over the smaller ring means it is finitelygenerated over the bigger ring. !By the structure theorem we now have:M∼=)C[T ]/(pi(T )ni)where pi(T ) is an irreducible polynomial. But C is algebraically closed.So, any polynomial has a root ai∈ C which means T − aidivides pi(T )which implies thatpi(T ) = T − aiup to multiplication by a scalar.Lemma 10.2. Let p(T ) = T − a. Then the p-primary part of M isnonzero (Mp'= 0) iff a is an eigenvalue of A.Proof. Suppose first that x ∈ M = Cnis an eigenvector of A witheigenvalue a. Then x '= 0 andAx = axwhich means thatpx = (A − a)x = 0.This implies that x is p-primary. So, x ∈ Mpmaking Mpnonzero.Conversely, suppose that Mp'= 0. Then there is an element x ∈ Mso that ann(x) = (pk). This meanspkx = (A − a)kx = 0andy = pk−1x = (A − a)k−1x '= 0But then(A − a)y = 0making y ∈ M into an eigenvector of A with eigenvalue a. !Since a is an eigenvalue, I started to write a = λ. Next I want tofind a basis for the cyclic module C[T ]/(pk) where p = T − λ.48 MATH 101A: ALGEBRA I PART B: RINGS AND MODULESLemma 10.3. If x ∈ Mpwith ann(x) = (pk) then the cyclic moduleRx = C[T ]x = C[A]x has basisy1= (A − λ)k−1x, y2= (A − λ)k−2x, ·, yk= (A − λ)0x = xas a vector space over C.Proof. The first point is to understand that if ann(x) = (pk) we meanthat pkis the minimal polynomial of A acting on x. In other words,f(A)x = 0 iff p(T )k|f(T ). This means two things.(1) (A − λ)kx = 0(2) If f(T ) is a nonzero polynomial of degree < k then f(A)x '= 0.The second statement can be reinterpretted as saying that(10.1) x, Ax, A2x, · · · , Ak−1xare linearly independent because any linear relation can be written ask*i=1ciAk−ix = 0.This says that f(A)x = 0 where f(T ) = c1Tk−1+· · ·+ckis a polynomialof degree k − 1 or less contradicting (2).When (1) is expanded out it says that Akx is a linear combinationof the vectors in (10.1). But then Ak+1is a linear combination ofAkx, Ak−1x, · · · , Axwhich is in the span of the vectors Ak−1x, · · · , x since Akis in thatspan. So, the vectors in (10.1) for a basis for the cyclic module Rx.This is not quite what we want. However, we can modify the basisabove in the following way. Suppose that fi(T ) is a polynomial ofdegree exactly equal to i and f0(T ) = c is a nonzero constant. Thenf0(A)x, f1(A)x, · · · , fk−1(A)xalso forms a basis for Rx since they have the same span (by inductionon k): the last vector fk−1(A)x is equal to a nonzero constant timesAk−1x plus lower terms. But the lower terms are in the span of Aix fori < k − 1 by induction. So, we get everything.Since fi(T ) = (T − λ)ihas degree i, we conclude that (T − λ)ix forma basis for R x as claimed. !10.3. the Jordan canonical form for A.Lemma 10.4. If ann(x) = (pk) where p = T − λ then the matrix ofA operating on the k-dimensional vector space Rx with respect to thebasis given by the above lemma is equal to the k × k Jordan block withdiagonal entries λ.MATH 101A: ALGEBRA I PART B: RINGS AND MODULES 49Proof. The basis elements yi= (A − λ)k−ix are related by the equationyi−1= (A − λ)yi.In other words,Ayi= λyi+ yi−1and Ay1= λy1. Since the jth column of the matrix for A is the vectorAyjwritten in terms of the basis y1, · · · , yk, the matrix isλ 1 0 · · · 00 λ 1 · · · 00 0 λ · · · 0...............0 0 0 · · · λThis is the Jordan block [λ]k. !Finally, I recalled how matrices behave with respect to change ofbasis. To make it really clear, let me first do the 2 × 2 case.Suppose that n = k = 2 and y1= (A − λ)x and y2= x. ThenAy1= λy1and Ay2= λy2+ y1. This can be written as the matrixformulaA(y1, y2) = (Ay1, Ay2) = (y1, y2)'λ 10 λ(If we write B = (y1, y2. · · · ) for the matrix with columns the basiselements y1, y2, etc. we getAB = BJwhere J is the Jordan matrix. The fundamental theorem for modulesover PID’s givesM∼=)C[T ]/((T − λi)ki)By the lemmas above, this implies the following.Theorem 10.5. After conjugating by a basis change matrix B, anyn × n complex matrix A is conjugate to a matrix in Jordan canonicalform:B−1AB = J =)[λi]ki.We need the uniqueness part of the fundamental theorem to tell usthat the pairs (λi, ki) are uniquely